Tony:
Have you ever thought about such a problem, the mass of A is measured when A is stationary, now A has a velocity of 0.8660254c, according to SR, the mass of A is no longer m, and the momentum of A is no longer mv , the kinetic energy of A is no longer 1/2mvv. What do you think? Is there really no need to consider SR here?
m is the rest mass. It does not change as an object moves faster.
You are correct that if you want to formally solve this problem using SR you will need to recognise that the initial momentum of A is $\frac{mv}{\sqrt{1-(v/c)^2}}$ and that the initial energy of A is $\frac{mc^2}{\sqrt{1-(v/c)^2}}$.
It is a fun exercise to work through the SR maths, though a lengthy one involving lots of algebra.
That's why I'm here, because I really don't know how to use SR to calculate this problem. If you can give an answer, you can share it here directly, you are always full of wisdom.
The correct answer is the same as your answer, even though you didn't do the maths correctly using SR.
After the collision, B has a velocity of 0.8660254c and A has zero velocity.
“The trick seems to be to use the centre of momentum frame of reference”, this method seems to greatly simplify the calculation, but when the reference frame changes, does the mass m of A and B also need to change?
No. m is the rest mass. It is the same in all frames of reference.
When A and B are stationary relative to the earth, we measured their mass as m, and now the reference system is no longer the earth, so is their mass still m?
Their rest masses don't change.
But from the analysis on the wiki, m1 and m2 have not changed due to the change of the reference system, which makes us wonder, is the earth the absolute reference system in the universe?
m1 and m2 are rest masses, which do not change.
Because SR has been proven to be the greatest and most correct theory in the world, obviously SR cannot be ignored.
How are you comparing all theories in the world? What is your measure of comparison?
James R gave a wrong conclusion again, and billvon also gave a wrong answer.
I agree with
your answer, Tony. You blundered into the correct answer even though you didn't use the correct theory to solve the problem. Call it a lucky accident!
Perhaps everyone has become accustomed to ignoring SR without hesitation when solving practical problems, and will tell you without hesitation that SR is the greatest theory that has been accurately verified when praising Einstein.
There's no need to ignore it. If you solve this particular problem with SR you get the same answer as the Newtonian/Galilean one. I did it, just for fun. You should check for yourself.
Can you solve problems using SR, Tony?
Taking the earth as the reference system, what are the masses of A and B before the collision?
You said they were both m. I assumed you meant that the rest masses were m. Real physicists don't use "relativistic mass", Tony.
What are the masses of A and B after the collision?
Their rest masses are still m. There's no need to introduce the idea of "relativistic mass" into the problem; it doesn't add anything useful.
What are the velocities of A and B after the collision?
Zero and 0.866c, respectively.
Please give the calculated data directly and a simple explanation.
Here's a reasonably simple explanation...
Consider the collision in the centre-of-mass frame of reference, which is initially moving at a speed of v/2 relative to the "ground". In that frame, the initial velocity of A is
almost v/2 and the initial velocity of B is -v/2. This is not
quite true, because velocity addition is a little more complicated in SR than it is in Newtonian/Galilean physics, but it's correct to within about 2%. (Note: if you do the calculation properly, the final answer to the problem is the same.)
Because the collision is elastic, momentum and total mechanical energy are conserved in the collision. The initial momentum of the two-mass system in the centre of mass frame is zero, so the final momentum must be zero in the same frame. The total energy in that frame is $2\gamma mc^2$, where $\gamma$ is the Lorentz factor corresponding to a speed of v/2 (i.e. the speed of each mass relative to the centre of mass).
Let's write down the momentum conservation equation in the centre-of-mass frame:
$$\gamma m\frac{v}{2} - \gamma m\frac{v}{2} = 0 = \gamma_1 mv_1 +\gamma_2 mv_2$$
And the energy conservation equation:
$$\gamma mc^2 + \gamma mc^2 = \gamma_1 mc^2 + \gamma_2 mc^2$$
In these equations, $m$ is the rest mass, $\gamma = \frac{1}{\sqrt{1-(v/c)^2}}$ and the other two Lorentz factors correspond to the (unknown) final velocities $v_1$ and $v_2$. For example $\gamma_1 \equiv \frac{1}{\sqrt{1-(v_1/c)^2}}$.
Now, all you have to do is to do the algebra and solve these two simultaneous equations to find the final velocities $v_1$ and $v_2$ in terms of the intial velocity $v$. The only physically viable solution is $v_1=-v/2, v_2=v/2$.
The final step is to convert the velocities back to the "ground" reference frame, where we find $v_1=0, v_2=v$.
(Note again: I have skimmed over the steps of correctly changing reference frames using the velocity addition formula from SR. But you should work through these steps yourself as an exercise and confirm that my answer is correct.)
So now you have to resort to asking irrelevant supplementary questions, in order to manufacture something to complain about. 